package dome1;

import java.util.*;

public class BinaryTree {
    static class TreeNode {
        public char val;
        public TreeNode left;//左孩子的引用
        public TreeNode right;//右孩子的引用

        public TreeNode(char val) {
            this.val = val;
        }
    }


    /**
     * 创建一棵二叉树 返回这棵树的根节点
     *
     * @return
     */
    public TreeNode createTree() {
        TreeNode A = new TreeNode('A');
        TreeNode B = new TreeNode('B');
        TreeNode C = new TreeNode('C');
        TreeNode D = new TreeNode('D');
        TreeNode E = new TreeNode('E');
        TreeNode F = new TreeNode('F');
        TreeNode G = new TreeNode('G');
        TreeNode H = new TreeNode('H');
        A.left = B;
        A.right = C;
        B.left = D;
        B.right = E;
        C.left = F;
        C.right = G;
        E.right = H;
        return A;
    }

    // 前序遍历
    public void preOrder(TreeNode root) {
        if (root == null) return;
        System.out.print(root.val + " ");
        preOrder(root.left);
        preOrder(root.right);
    }

    // 中序遍历
    void inOrder(TreeNode root) {
        if (root == null) return;
        inOrder(root.left);
        System.out.print(root.val + " ");
        inOrder(root.right);
    }

    // 后序遍历
    void postOrder(TreeNode root) {
        if (root == null) return;
        postOrder(root.left);
        postOrder(root.right);
        System.out.print(root.val + " ");
    }

    public static int nodeSize;

    /**
     * 获取树中节点的个数：遍历思路
     */
    void size(TreeNode root) {
        if (root == null) return;
        nodeSize++;
        size(root.left);
        size(root.right);
    }

    /**
     * 获取节点的个数：子问题的思路
     *
     * @param root
     * @return
     */
    int size2(TreeNode root) {
        if (root == null) return 0;
        return size2(root.left) + size2(root.right) + 1;
    }


    /*
     获取叶子节点的个数：遍历思路
     */
    public static int leafSize = 0;

    void getLeafNodeCount1(TreeNode root) {
        if (root == null) return;
        if (root.left == null && root.right == null) leafSize++;
        getLeafNodeCount1((root.left));
        getLeafNodeCount1((root.right));
    }

    /*
     获取叶子节点的个数：子问题
     */
    int getLeafNodeCount2(TreeNode root) {
        if (root == null) return 0;
        if (root.left == null && root.right == null) return 1;
        return getLeafNodeCount2(root.left) + getLeafNodeCount2(root.right);
    }

    /*
    获取第K层节点的个数
     */
    int getKLevelNodeCount(TreeNode root, int k) {
        if (root == null) return 0;
        if (k == 1) return 1;
        int left = getKLevelNodeCount(root.left, k - 1);
        int right = getKLevelNodeCount(root.right, k - 1);
        return left + right;
    }

    /*
     获取二叉树的高度
     时间复杂度：O(N)
     */
    int getHeight(TreeNode root) {
        if (root == null) return 0;
        if (root.left == null && root.right == null) return 1;
        int left = getHeight(root.left);
        int right = getHeight(root.right);
        return Math.max(left, right) + 1;
    }


    // 检测值为value的元素是否存在
    TreeNode find(TreeNode root, char val) {
        if (root == null) return null;
        if (root.val == val) return root;
        TreeNode leftNode = find(root.left, val);
        if (leftNode != null) return leftNode;
        TreeNode rightNode = find(root.right, val);
        if (rightNode != null) return rightNode;
        return null;
    }

    //层序遍历
    void levelOrder(TreeNode root) {
        if (root == null) return;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            TreeNode cur = queue.poll();
            System.out.print(cur.val + " ");
            if (cur.left != null) queue.offer(cur.left);
            if (cur.right != null) queue.offer(cur.right);
        }
    }


    // 判断一棵树是不是完全二叉树
    boolean isCompleteTree(TreeNode root) {
        if (root == null) return true;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            TreeNode cur = queue.poll();
            if (cur != null) {
                queue.offer(cur.left);
                queue.offer(cur.right);
            } else {
                break;
            }
        }
        while (!queue.isEmpty()) {
            TreeNode cur = queue.peek();
            if (cur != null) {
                return false;
            } else {
                queue.poll();
            }
        }
        return true;
    }

    public List<Character> postorderTraversal(TreeNode root) {
        List<Character> list = new ArrayList<>();
        if(root == null) return list;
        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = root;
        TreeNode prev = null;
        while(cur!=null || !stack.isEmpty()) {
            while(cur!=null) {
                stack.push(cur);
                cur = cur.left;
            }
            TreeNode top = stack.peek();
            if(top.right==null || top.right==prev) {
                list.add(top.val);
                stack.pop();
                prev = top;
            }else {
                cur = cur.right;
            }
        }
        return list;
    }
}